Certification Problem

Input (TPDB TRS_Standard/AG01/#3.1)

The rewrite relation of the following TRS is considered.

minus(x,0) x (1)
minus(s(x),s(y)) minus(x,y) (2)
quot(0,s(y)) 0 (3)
quot(s(x),s(y)) s(quot(minus(x,y),s(y))) (4)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Switch to Innermost Termination

The TRS is overlay and locally confluent:

10

Hence, it suffices to show innermost termination in the following.

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
minus#(s(x),s(y)) minus#(x,y) (5)
quot#(s(x),s(y)) quot#(minus(x,y),s(y)) (6)
quot#(s(x),s(y)) minus#(x,y) (7)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.